Zauważ, że: [latex]frac{n+3}{n+1}=frac{n}{n+1}+frac{3}{n+1}=frac{n+1}{n+1}-frac{1}{n+1}+frac{3}{n+1}=1+frac{2}{n+1} [/latex] I teraz: [latex]lim_{n o infty} (frac{n+3}{n+1})^{2n+1}=lim_{n o infty} (1+frac{2}{n+1})^{2n+1}= \ =lim_{n o infty} left( left( left( 1+frac{2}{n+1} ight) ^{frac{n+1}{2}} ight)^{frac{2}{n+1}} ight)^{2n+1}=lim_{n o infty} e^{frac{4n+2}{n+1}}=\ =lim_{n o infty} e^{frac{4+frac{2}{n}}{1+frac{1}{n}}}=e^{4}[/latex] W drugim zauważ, że: [latex]frac{n+2}{n-4}=frac{n}{n-4}+frac{2}{n-4}=frac{n-4}{n-4}+frac{4}{n-4}+frac{2}{n-4}=1+frac{6}{n-4}[/latex] A zatem: [latex]lim_{n o infty} (frac{n+2}{n-4})^{n}=lim_{n o infty} (1+frac{6}{n-4})^{n} = \ =lim_{n o infty} left( left( left( 1+frac{6}{n-4} ight)^{frac{n-4}{6}} ight)^{frac{6}{n-4}} ight)^{n}=lim_{n o infty} e^{frac{6n}{n-4}}= \ =lim_{n o infty} e^{frac{6}{1-frac{4}{n}}}=e^{6}[/latex] Wykorzystujemy oczywiście wzór: [latex]lim_{n o infty} left( 1+frac{1}{n} ight) ^{n}=e[/latex]
